Re: [ippm] draft-ietf-ippm-rfc8889bis

[John Scudder <[e-mail]>]

Yes, that’s a lot clearer, thanks.

But, although it’s clearer, it brings me back to the question that took me to the document to begin with — doesn’t that statement describe any connected graph of routers, from a single router all the way up to an entire network? In order to drain or introduce packets you have to include a host, since you’ve already excluded packet loss. (I’m assuming that you’re excluding packets that are destined for routers themselves, e.g. management traffic.)

—John

> On Jul 6, 2022, at 3:42 AM, Giuseppe Fioccola <[e-mail]> wrote:
> 
> 
> Hi John,
> Thank you for raising this point.
> A cluster is a subnetwork where the number of packets that go in is the same as the number that go out, as also stated in the Terminology section.
> 
> We can modify the wording in Section 5 accordingly:
> 
> "A cluster graph is a subnetwork of the entire monitoring network graph that still satisfies the condition that the number of packets that go in is the same as the number that go out, if no packet loss occurs."
> 
> Regards,
> 
> Giuseppe
> 
> -----Original Message-----
> From: John Scudder <[e-mail]>
> Sent: Tuesday, July 5, 2022 11:05 PM
> To: [e-mail]; [e-mail]
> Subject: draft-ietf-ippm-rfc8889bis
> 
> Hi Authors (and WG),
> 
> I have a question about Section 5, Network Clustering. It says,
> 
>   A cluster graph is a subnetwork of the entire monitoring network
>   graph that still satisfies the packet loss equation (introduced in
>   the previous section), where PL in this case is the number of packets
>   lost in the cluster.  As for the entire monitoring network graph, the
>   cluster is defined on a per-flow basis.
> 
> But when I look back at the previous section, all I see is
> 
>   PL = (PI1 + PI2 +...+ PIn) - (PO1 + PO2 +...+ POm)
> 
> But that’s just a definition of what PL is, it can take on any value based on its inputs. What does it mean for it b be “satisfied”? That’s a term that typically requires us to state what condition is to be satisfied, and as far as I can tell you haven’t stated a condition.
> 
> Thanks,
> 
> —John